klastiesym

Answered

2022-10-24

Find the Laplace transform of the function $H(t-a){t}^{n}$. That's what I have done so far:

$$L\{H(t-a){t}^{n}\}={\int}_{0}^{\mathrm{\infty}}{e}^{-st}H(t-a){t}^{n}dt={\int}_{a}^{\mathrm{\infty}}{e}^{-st}{t}^{n}dt={\int}_{0}^{\mathrm{\infty}}{e}^{-s(t+a)}(t+a{)}^{n}dt$$

Could you tell me how I can continue?

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Answer & Explanation

t5an1izqp

Expert

2022-10-25Added 13 answers

You can obtain directly from the integral

$$\mathcal{L}\{H(t-a){t}^{n}\}={\int}_{a}^{\mathrm{\infty}}{e}^{-st}{t}^{n}dt=\frac{1}{{s}^{n+1}}{\int}_{as}^{\mathrm{\infty}}{e}^{-z}{z}^{n}dz=\frac{1}{{s}^{n+1}}\mathrm{\Gamma}(n+1,as)$$

recalling the integral expression of incomplete Gamma function for integer n

$$\mathrm{\Gamma}(n,x)={\int}_{x}^{\mathrm{\infty}}{e}^{-z}{z}^{n-1}dz.$$

NOTE Another useful methodology may be the following.

Observing that

$$H(t-a){t}^{n}=H(t-a)(t-a+a{)}^{n}=H(t-a)\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}(t-a{)}^{k}{a}^{n-k}$$

the Laplace transform becomes

$$\mathcal{L}\{H(t-a){t}^{n}\}=\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{a}^{n-k}\mathcal{L}\{H(t-a)(t-a{)}^{k}\}$$

recalling that $\mathcal{L}\{H(t-a)f(t-a)\}={e}^{-as}F(s)$. Using $\mathcal{L}\{{t}^{n}\}=\frac{n!}{{s}^{n+1}}$, we obtain

$$\mathcal{L}\{H(t-a){t}^{n}\}={e}^{-as}\sum _{k=0}^{n}{\textstyle (}\genfrac{}{}{0ex}{}{n}{k}{\textstyle )}{a}^{n-k}\frac{k!}{{s}^{k+1}}=\frac{n!{e}^{-as}}{{s}^{n+1}}\sum _{k=0}^{n}\frac{(as{)}^{n-k}}{(n-k)!}=\frac{1}{{s}^{n+1}}n!{e}^{-as}\sum _{\nu =0}^{n}\frac{(as{)}^{\nu}}{\nu !}$$

and recalling that

$$\mathrm{\Gamma}(n,z)=(n-1)!{e}^{-z}\sum _{\nu =0}^{n-1}\frac{{z}^{\nu}}{\nu !}$$

is the incomplete Gamma function for integer n, we obtain finally

$$\mathcal{L}\{H(t-a){t}^{n}\}=\frac{1}{{s}^{n+1}}\mathrm{\Gamma}(n+1,as).$$

Pellagra3d

Expert

2022-10-26Added 1 answers

You know that $\mathcal{L}\{\mathcal{U}(t-a)\}=\frac{{e}^{-as}}{s},a>0$. And

$$\mathcal{L}\{{t}^{n}f(t)\}=(-1{)}^{n}{\textstyle (}\mathcal{L}(f(t)){{\textstyle )}}^{(n)}$$

Now let think about $(}\frac{{e}^{-as}}{s}{{\textstyle )}}^{(n)$. Using Maple, I found:

$$n=1\to {\left(\frac{{e}^{-as}}{s}\right)}^{\prime}=-\mathrm{exp}(-as)\frac{as+1}{{s}^{2}}\phantom{\rule{0ex}{0ex}}n=2\to {\left(\frac{{e}^{-as}}{s}\right)}^{\u2033}=\mathrm{exp}(-as)\frac{{a}^{2}{s}^{2}+2as+2}{{s}^{3}}\phantom{\rule{0ex}{0ex}}n=3\to {\left(\frac{{e}^{-as}}{s}\right)}^{\u2034}=-\mathrm{exp}(-as)\frac{{a}^{3}{s}^{3}+3{a}^{2}{s}^{2}+6as+6}{{s}^{4}}$$

And so I think the possible pattern could be

$$(-1{)}^{n}\frac{\sum _{k=0}^{n}\frac{n!}{(n-k)!}{s}^{n-k}{a}^{n-k}}{{s}^{n+1}}$$

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